Problem 18-1 Repair calls are handled by one repairman at a photocopy shop. Repa
ID: 336436 • Letter: P
Question
Problem 18-1 Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed, witha mean of 1.9 hours per call. Requests for copier repairs come in at a mean rate of 2.4 per eight-hour day (assume Poisson) a. Determine the average number of customers awaiting repairs. (Round your answer to 2 decimal places.) Number of customers b. Determine system utilization. (Round your answer to 2 decimal places. Omit the "%" sign in your response.) System utilization % c. Determine the amount of time during an eight-hour day that the repairman is not out on a call. (Round your answer to 2 decimal places.) Amount of timehours d. Determine the probability of two or more customers in the system. (Do not round intermediate calculations. Round your answer to 4 decimal places.) Probability [Explanation / Answer
This is M/M/1 queueing model
Arrival rate, a = 2.4 / 8 = 0.3 per hour
Service rate, s = 1/1.9 = 0.5263 per hour
a) Average number of customers awaiting repairs (Lq) = a2/(s*(s-a)) = 0.32/(0.5263*(0.5263-0.3)) = 0.76 (this is the number of customers who are yet to be visited by the repairman)
b) System utilization, u = a/s = 0.3/0.5263 = 57 %
c) Amount of time = 8*(1-57%) = 3.44 hours
d) Probability of n customers in system, Pn = (1-u)un
P0 = (1-0.57)*0.570 = 0.43
P1 = (1-0.57)*0.571 = 0.2451
Probability of 2 or more customers in system = 1 - (P0 + P1) = 1 - (0.43 + 0.2451) = 0.3249
Probability = 0.3249